Question

In a certain decay process, a nucleus in the vicinity of mass 240 emits alpha particles with the following energies (in MeV): 5.545 ( ao), 5.513 ( al), 5.486( a2), 5.469 ( a3), 5.443 (a4), 5.417 (a5), and 5.389 (a6). The following gamma rays in the daughter nucleus are seen (energies in keV): 26 (g1), 33(g2), 43(g3), 56(g4), 60(g5), 99(g6), 103(g7), 125(g8). Construct a decay scheme from this information, assuming a0. populates the ground state of the daughter. Keep in mind that energies can have uncertainty in measurements around 1-2 keV.

Answer 1

the energy of the alpha decay is not the same as the Q-value, be-
cause the 237Np recoil nucleus also takes away some of that kinetic energy released by
the Q-value. One can use conservation of momentum to find out the corresponding
Q-value for each alpha energy:
pα = pNp (3)
mαva = mNpvNp (4)
2mαEα = 2mNpENp (5)
mα ENp = Eα (6) mNp
We know the energy of the alpha particle in each case, and using this formula, we
can find the corresponding Q-value for each alpha energy:
mα mα 4.0026032 amu Q = Eα + ENp = Eα + Eα = Eα 1 + = Eα 1 + ≈ 1.016885Eα mNp mNp 237.0481673 amu
(7)
This yields the following table of Q-values for our three alpha particle energies, and
the corresponding 237Np excited states by subtracting these Q-values from the 241Am
relative ground state energy (5637.8 keV):
Eα (keV) Q (keV) 237Np Energy Level (Calc.) Closest 237Np Level (Table)
5485.56 5578.18 59.62 59.5
5442.80 5534.70 103.1 103.0
5388.23 5479.21 158.59 158.5
Clearly the values are remarkably close. The full decay diagram proceeds as follows:
One can see that not every energy transition is allowed. The following transitions
were identified, along with their corresponding gamma ray energies. In addition,
allowed electron transitions following internal conversion are tabulated for just the
first four energy levels:
Ei Ef ΔE K-eject?
√
L-eject?
√ 158.5 33.2 125.3
√ 158.5 59.5 99.0
√ 158.5 103.0 55.5
√ 130.0 33.2 96.8
√ 130.0 75.9 54.1
√ 103.0 0 103.0
√ 103.0 33.2 69.8
√ 103.0 59.5 43.5
√ 103.0 75.9 27.1
√ 75.9 0 75.9
√ 75.9 33.2 42.7
√ 59.5 0 59.5
√ 59.5 33.2 26.3
√ 33.2 0 33.2
(b) For each of the initial alpha particle energies, separately sketch a hypothetical photon (gamma
plus x-ray) spectrum that you would expect to observe. You may want to use the NIST X-Ray
Transition Energy Database to help generate your answer.
This spectrum should consist of all the gamma ray energy differences in the table
above, along with all the x-ray transition energies identified, as at least one IT/IC
process is energetic enough to eject a K-shell electron. The spectrum would therefore
look something like this:
3. It is clear that 241Am produces a few types of radiation at many different energies. Do you expect the
alpha particles, the gamma rays, or the x-rays to be responsible for producing the largest number of
ions in a fixed space (like in a smoke detector), and why?
The alpha particles should be responsible for most of the ionizations, because they interact
so much more strongly with matter, and therefore the electron clouds in the matter. Their
heavy mass and high charge compared to gammas or x-rays (no mass, no charge) gives
them far higher ionizing power.
Medical Isotope Physics
In these problems, consider the decay of 99Mo, a crucial medical isotope widely used in imaging and diagnosis
procedures.
1. Calculate the Q-value for the decay of 99Mo using the binding energies of the initial and final nuclei,
and any other information that you need.
The decay of 99Mo proceeds by beta decay to 99mTc, followed by a gamma decay (IT) to stable 99Tc:
99Mo →99m T c + β− + ν¯99mT c →99 T c + γ (8)
All that is required are the masses in amu of 99Mo and 99Tc, and the conversion factor between amu
and MeV:
2 2 931.49 MeV Q = mMo−99zc − mT c−99zc = 931.49 (98.9077116 − 98.9062546) = 1.3572 MeV − z2 amu c
(9)
This is indeed the difference in energy levels according to the decay diagram for
You may have noticed that 99Mo is an unstable isotope. Which nuclear reactions could create 99Mo?
Write the nuclear reactions for these processes, and calculate their Q-values to justify your answer.
This is where some creativity can come into play. 99Mo could either be produced by
spontaneous decay reactions, or by deliberate bombardment of parent isotopes with other
particles. First, the decay modes:
103 2 931.49 MeV Ru →99 Mo + α; Q = (mRu−103 − mMo−99 − mα)zc = −3.71 MeV (10)
amu −z2c
99N b →99 Mo + β− + ν¯; Q = 1.3572 MeV (above) (11)
99T c →99 Mo + β+ + ν; Q = −1.3572 MeV (12)
99T c →99 Mo + e− (EC) ; Q = −1.3572 MeV (13)
100Mo →99 Mo + n; Q = −8.29 MeV (14)
235U →99 Mo +134 Sn + 21
0n; Q = 177.4 MeV (15)
Now for the energetic particle bombardment methods:
98N b + p+ →99 Mo; Q = 9.73 MeV (16)
98Mo + n →99 Mo; Q = 5.93 MeV (17)
95Zr + α →99 Mo; Q = 2.73 MeV (18)
As one can see, only beta decay and spontaneous fission is possible, though a number of
bombardment options are available.
3. For the most common 99Mo production method which does not arise from spontaneous radioactive
decay, answer the following questions:
(a) Assuming the incoming particle was at room temperature to begin with (you should find its kinetic
energy), what are the possible recoil kinetic energies of the 99Mo produced?
The most likely and available method to produce 99Mo is by neutron capture in a
reactor, as this is the most readily available source of particles already being created.
The recoil question is a bit of a trick question, as only one particle (99Mo) is created,
from the merging of two other particles with about zero kinetic energy or momentum.
Therefore, the 99Mo nucleus should still stay at rest, so it should have zero recoil
energy. This is because 99Mo is in effect a long-lived compound nucleus, which then
decays by beta decay to 99Tc. The Q-value in this reaction is absorbed into creating
this unstable nucleus.
(Not required for credit, just good to know): If one wants to explore deeper, the
full reaction involves creating this excited compound nucleus of 99Mo, which then
de-excites to the ground state of 99Mo via isomeric transition. One could then equate
the momentum of the 99Mo nucleus and the outgoing gamma ray:
E2 Eγ γ 2mMo−99EMo−99 = ; EMo−99 = (19) c 2mMo−99c2
(b) How will the recoil energy and the outgoing radiation energy change if the incoming radiation
had a kinetic energy of 2 MeV?
If the neutron had an incoming kinetic energy of 2 MeV, then the momentum of the
excited compound nucleus should be equal to that of the incoming neutron:
mn 2mnEn = 2mMo−99EMo−99; EMo−99 = En (20) mMo−99